The atomic mass unit, or amu, is 1/12 the mass of one atom of carbon-12. An atomic mass unit has the mass of 1/(6.0221415 * 10^23) grams.
The charge of the ion is +1
Answer:
- Part a) 0.0104 moles copper(II) nitrate.
i) 0.0418 mole Cu
ii) 0.0209 mol Ag NO₃
Explanation:
<u>1) Balanced chemical reaction (single replacement):</u>
In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)
- Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag
<u>2) Mole ratio: </u>
- 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag
<u />
<u>3) Moles of Ag</u>
- n = mass in grams / atomic mass
- atomic mass of Ag: 107.868 g/mol
- n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag
<u>4) Moles of copper(II) nitrate:</u>
- Set the proportion using the mole ratio:
- 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
- Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ = 0.0104 moles Cu(NO₃)₂
That is the answer of part a: 0.0104 moles copper(II) nitrate.
<u>5) Moles of each reactant</u>
i) Cu:
- Set a proportion using the theoretical mole ratio
1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag
- Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu
ii) Ag NO₃
- Set a proportion using the teoretical mole ratio
2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag
- Solve for x: x = 0.0209 mol Ag NO₃
Answer:
0.043 M
Explanation:
The reaction that takes place is:
- Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume required to reach the equivalence point</em>:
- 0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺
As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.
With that in mind we can <u>calculate the hydroxide ion concentration in the original sample solution</u>, using <em>the calculated number of moles and given volume</em>:
- 1.0817 mmol OH⁻ / 25.0 mL = 0.043 M
<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O
2C4H10 + 13 O2 ----> 8CO2 + 10H2O
1. Count the C on the left (4), put a 4 where the C on the right.
2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O
3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left.
4. Now multiply everything on the equation by two so you have nice integer numbers.
5. check you have the same amount of everything on each side.
Example C: left 8, right 8, etc.
I hope this helps. :)</span><span>
</span>
